A definite amount of NH4cl was boiled with 100ml of 0.8N naoh for complete reaction. After the reaction, the reactant mixture containing excess of naoh was neutralized with 12.5ml of 0.75N nh2s04. Calculate the amount of nh4cl taken.
The answer 3.78g
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# milliequivalents = 100 x 0.8 = 80 total
-# m.e. backtitrated with NaoH = 12.5 x 0.75 = 9.375
NH4Cl initially = 80-9.375= 70.625 then
70.625 x m.e.w NH4Cl (about 53.5) = abpit 3.8 g.
To calculate the amount of NH4Cl taken, we need to use the concept of Stoichiometry, which involves balancing the chemical equation and using the given information to determine the number of moles involved.
Let's start by writing the balanced chemical equation for the reaction between NH4Cl and NaOH:
NH4Cl + NaOH -> NaCl + H2O + NH3
According to the equation, 1 mole of NH4Cl reacts with 1 mole of NaOH.
Now, let's calculate the number of moles of NaOH used in the reaction:
Moles of NaOH = Volume of NaOH solution (in liters) x Normality of NaOH
= 100 ml (or 0.1 L) x 0.8 N
= 0.08 moles
Next, let's determine the number of moles of H2SO4 needed to neutralize the excess NaOH:
Moles of H2SO4 = Volume of H2SO4 solution (in liters) x Normality of H2SO4
= 12.5 ml (or 0.0125 L) x 0.75 N
= 0.009375 moles
Since NaOH and H2SO4 react in a 1:2 ratio, we can say that the moles of NaOH consumed in the reaction with NH4Cl are double the moles of H2SO4 used in the neutralization.
Thus, moles of NH4Cl = 2 x Moles of H2SO4
= 2 x 0.009375
= 0.01875 moles
Finally, we can calculate the mass of NH4Cl using its molar mass (53.49 g/mol):
Mass of NH4Cl = Moles of NH4Cl x Molar mass of NH4Cl
= 0.01875 moles x 53.49 g/mol
≈ 1.00 g
Therefore, the amount of NH4Cl taken is approximately 1.00 g.
However, the given answer is 3.78 g, which suggests a different solution approach or a possible error in the given information. Please verify the question or review the solution steps to reconcile the discrepancy.